See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Selective deprotonation and alkylation of a molecule bearing two different OH groups — a phenolic OH and an aliphatic (secondary alcohol) OH — using one equivalent of base followed by an alkylating agent. Step 1 – Identify the two OH groups in the starting material: • A phenolic OH (ArOH) attached directly to the aromatic ring: pKa ≈ 10 • An aliphatic secondary alcohol OH on the seven-membered ring: pKa ≈ 16–17 Step 2 – Effect of one equivalent of NaOH: NaOH (pKa of water ≈ 15.7) is a strong enough base to deprotonate the more acidic phenolic OH (pKa ~10) selectively and quantitatively, but it is NOT strong enough to deprotonate the aliphatic alcohol (pKa ~16–17) to any significant extent. Therefore, with exactly one equivalent of NaOH, only the phenoxide anion is formed; the aliphatic OH remains protonated. Step 3 – Reaction with MeBr (methyl bromide, an SN2 alkylating agent): The phenoxide anion, being a good nucleophile, reacts with MeBr via SN2 to give the aryl methyl ether (ArOMe). The aliphatic OH is unreacted and remains as a free OH. Step 4 – Structure of product (A): The product has OMe at the aromatic ring position and a free OH on the aliphatic seven-membered ring. This matches option (a). Why other options fail: • Option (b): This would require selective methylation of the aliphatic OH while leaving the phenolic OH free, which is the opposite of what happens since the phenol is far more acidic and is preferentially deprotonated by NaOH. • Option (c): MeBr under these mild conditions does not cause elimination to form a double bond; there is no mechanism for ring dehydration here. • Option (d): 'None of these' is incorrect because option (a) correctly describes the product. Therefore, the correct answer is A.